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tanad
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6t=4*27
t=18 hours
27-18=9 fewer hours
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tanad
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?

A: 9
B: 12
C: 16
D: 18
E: 24

This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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tanad
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?

A: 9
B: 12
C: 16
D: 18
E: 24

Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours

Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
Answer A
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given 4 machines can complete the job in 27 hrs,
--> rate of 4 machines = 1/27
--> rate of each machine = (1/27)/4
--> rate of 6 machines = 6 * ( (1/27)/4 )= 1/18
hence , time taken by 6 machines = 18 hrs
difference = 27 - 18 = 9
option A
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Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?

A: 9
B: 12
C: 16
D: 18
E: 24

Total work = 27 * 4

Time taken when 6 machines work = \(\frac{(27*4)}{6}\) => 18 hours

So, working together 6 machines take 9 hours less ( 27 - 18 )

Answer will be (A)

Quote:


Bunuel Sir and Abhishek Sir,
Please tell me if my approach is wrong.

Let "r" be the rate of the machine.

Rate * Time = Workdone
So,

4 r * 27 = 6 r *X

>>> X = 4*27/6
>>> X = 18

How many fewer hours does it take = 27 -18 = 9

Please let me know.
Thanks again!

Regards,
Yosita

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1/(4*27)=1/108=rate of 1 machine
let t=time
t*6*(1/108)=1
t=18
27-18=9
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Time taken my 4 machines= 27 hours.

As the efficiencies of the machines are equal, we can take the time taken by one machine and multiply it by the time taken 4 machines we get the total hours taken by all of them.

So we can find the time taken by 1 machine= 27*4=108 hours. That is basically the time effort of 4 machines will be required by the single machine to complete the work.

6 machines will require= 108/6=18 hours.

More time required by 6 machines=27-18=9 hours.
A
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We can use chain formula here-
P1*D1*H1= P2*D2*H2 where, P D H are person, days and hours respectively.
6*x= 4*27 , ignoring the D since no days are given.
solving for x, we get x=18 hours.

But STOP here and re read the Q again. It says how much LESS time, not how much TIME.
Hence less time = 27-18=9 hours
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tanad
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?

A: 9
B: 12
C: 16
D: 18
E: 24

If 4 machines take a total of 27 hours to complete a production order, then 1 machine will take \(27*4 = 108\) hours to complete a production order.

Therefore, 6 machines will take \(\frac{108}{6} = 18\) hours.

Difference in number of hours taken by 6 machines vs 4 machines in fulfilling a production order = \(27-18 = 9\) hours.

Answer choice (A) is correct, IMO.
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I get the correct answer, when I find the speed of 6 machines by doing 6/27*4 = 1/18, hence 18 hours, and then minus 9 hours.
However, if I directly do:
6/(27*4)-(1/27), it produces = 1/54, or 54 hours. I wonder, what is fundamentally the error with the second approach?

KarishmaB
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TargetMBA007
I get the correct answer, when I find the speed of 6 machines by doing 6/27*4 = 1/18, hence 18 hours, and then minus 9 hours.
However, if I directly do:
6/(27*4)-(1/27), it produces = 1/54, or 54 hours. I wonder, what is fundamentally the error with the second approach?

KarishmaB

Recall that we can easily solve these questions using the concept of Joint variation:

4.................27
6................. ??

T = 27 * (4/6)
because time taken will reduce when number of machines increases

T = 18 i.e. time taken will be 9 hrs less

In your second approach, you are calculating 'Rate of 6 machines - Rate of 4 machines' which gives you 'Rate of 2 machines' since rates are additive.
(you know Rate of 4 machines + Rate of 2 machines = Rate of 6 machines)

So 1/54 is the rate of 2 machines and 54 hrs is the time taken when only 2 machines are working. What we are actually asked is the simple subtraction 'time taken by 4 machines - time taken by 6 machines'
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